What is the integral of 1/(sqrt(x)(9x+4)) ?

The integral of 1/(sqrt(x)(9x+4)) is 1/3 arctan(3/2 sqrt(x))+C

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integral of 1/(sqrt(x)(9x+4))

\int \frac{1}{x ( 9 x + 4 )} d x

\frac{1}{3} arctan (\frac{3}{2} x) + C

Solution steps

\int \frac{1}{x ( 9 x + 4 )} d x

Apply u-substitution

: \int \frac{2}{9 u ^{2} + 4} d u

= \int \frac{2}{9 u ^{2} + 4} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{1}{9 u ^{2} + 4} d u

Apply Integral Substitution

= 2 \cdot \int \frac{1}{6 ( v ^{2} + 1 )} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \frac{1}{6} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= 2 \cdot \frac{1}{6} arctan (v)

Substitute back

= 2 \cdot \frac{1}{6} arctan (\frac{3}{2} x)

Simplify

2 \cdot \frac{1}{6} arctan (\frac{3}{2} x) : \frac{1}{3} arctan (\frac{3}{2} x)

= \frac{1}{3} arctan (\frac{3}{2} x)

Add a constant to the solution

= \frac{1}{3} arctan (\frac{3}{2} x) + C

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What is the integral of 1/(sqrt(x)(9x+4)) ?
The integral of 1/(sqrt(x)(9x+4)) is 1/3 arctan(3/2 sqrt(x))+C